Question: Graph this system of equations and solve. $y = -\dfrac{5}{2} x + 4$ $2x-2y = 6$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $-\dfrac{5}{2}$ . Remember that the slope tells you rise over run. So in this case for every $5$ positions you move down (because it's negative) You must also move $2$ positions to the right. $2$ positions to the right. $5$ positions down from $(0, 4)$ is $(2, -1)$ Graph the blue line so it passes through $(0, 4)$ and $(2, -1)$ Convert the second equation, $2x-2y = 6$ , to slope-intercept form. $y = x - 3$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ positions to the right. $1$ position to the right. $1$ position up from $(0, -3)$ is $(1, -2)$ Graph the green line so it passes through $(0, -3)$ and $(1, -2)$ The solution is the point where the two lines intersect. The lines intersect at $(2, -1)$.